Pictures of Julia and Mandelbrot Sets/Julia and Mandelbrot sets for transcendental functions

For a transcendental complex function, such as sin(z),cos(z),exp(z),...{\displaystyle sin(z),cos(z),exp(z),...}, which must be assigned degree ∞ and which has ∞ as an attracting fixed point, the potential function for the Fatou domain containing ∞ does not exist, and therefore the colouring cannot be made smooth in the usual way. Besides this, it is possible that the status of ∞ as an attracting fixed point is ambiguous.

Julia set for sin(z)+c{\displaystyle sin(z)+c}

In the Sea Horse Valley of a mini-mandelbrot of the Mandelbrot set for sin(z)+c{\displaystyle sin(z)+c}

This is the case for sin(z){\displaystyle sin(z)} and cos(z){\displaystyle cos(z)}. Sin(z){\displaystyle Sin(z)} can be defined by sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(y){\displaystyle sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(y)}, and we see from this formula, that if we go towards ∞ along a vertical line, the value grows (exponentially) to ∞, but if we go towards ∞ along a horizontal line, the value remains bounded. As an iteration of z by sin(z)+c{\displaystyle sin(z)+c} can be small when z has an arbitrarily high y-value (namely if cos(x){\displaystyle cos(x)} is near 0), the inner Fatou domains extend towards ∞ in the vertical direction, and also in the horizontal direction, because of the periodicity. The same applies therefore for the Julia set. The Fatou domain containing ∞ must here be defined as the Fatou domain containing points having arbitrarily large y-values, but this Fatou domain is not an open set: it has no interior points. In the colouring it is therefore inseparable from the Julia set, which consists of infinitely dense lying threads. So, if there are no inner Fatou domains, the Julia set is lying densely in the plane, implying that the whole plane should be coloured as the boundary. Nevertheless, the computer gives us a non-trivial picture (top picture).

The reason is that we are forced to use a relatively small radius of the large circle determining the stopping of the iteration, owing to the exponential growth of sin(z){\displaystyle sin(z)} in the y-direction. Therefore the sequences of iteration stop after only few iterations, and we colour on the basis of the number of iterations. As the colour of a point c outside the Mandelbrot set is the colour of the (second) critical point of the Fatou domain for c containing ∞, the domain outside the Mandelbrot set is, like the outer Fatou domain, without interior points: it is interwoven with infinitely lying threads. This wire mesh makes up a continuation of the usual boundary, which is unaffected by the phenomenon, as the distance function is unaffected by the nature of the function. For a rational function, the boundary consists of the points such that the associated Julia set contains the (second) critical point. However, for a transcendental function this set can be larger than the boundary constructed from the distance function, and in our case it lies densely in the domain outside the interior of the Mandelbrot set. Nevertheless we get a colouring, because the iterations stop early. We are here in the Sea Horse Valley of a mini-mandelbrot of the Mandelbrot set for sin(z)+c{\displaystyle sin(z)+c} (middle picture).

For iteration towards finite cycles, the Julia sets look like those for rational functions. But it can happen that there are small circles in the picture of only one colour, because it is impossible here, at a specific step in the iteration, to calculate the next value of the transcendental function in the formula.

The Mandelbrot set for (1−z2)/(z−z2cos(z))+c{\displaystyle (1-z^{2})/(z-z^{2}cos(z))+c} has a look that is typical for the rational functions where the iterations are towards finite cycles (first picture below). This Mandelbrot set is constructed from two conjugate critical points, and if we let z1 and z2 be the images of these by (1−z2)/(z−z2cos(z)){\displaystyle (1-z^{2})/(z-z^{2}cos(z))}, we have that if c is real and numerical large and belonging to the Mandelbrot set, then z1 + c and z2 + c belong to different Fatou domains for (1−z2)/(z−z2cos(z))+c{\displaystyle (1-z^{2})/(z-z^{2}cos(z))+c}, therefore 1/cos(z1+c)+c{\displaystyle 1/cos(z1+c)+c} and 1/cos(z2+c)+c{\displaystyle 1/cos(z2+c)+c} belong to different Fatou domains, and if we add a multiple of 2π{\displaystyle 2\pi } to c (so that the result still is numerically large), then these two numbers will belong to different Fatou domains. Therefore we can say: it is possible that the Mandelbrot set extends towards infinity in the horizontal direction. It seems really to be the case, as we can see if we draw the Mandelbrot set and zoom out.

The two following pictures show sections of the Julia set for c = -1. The point z = -1 is a fixed point for this iteration, and as the derivative of the function in this point is numerically larger than 1, z = -1 is repelling and belongs therefore to the Julia set. If x is real and numerically very large, the iteration of x is near 1/cos(x)−1{\displaystyle 1/cos(x)-1}. Therefore, if x* is a real point of the Julia set such that x* - (-1) > 1, we can find a real point of large numerical value that iterates into this point of the Julia set, and this indicates that the Julia set extends towards infinity in the horizontal direction. And as cos(z) grows exponentially in the vertical direction, the iteration of a point z of large numerical y-value is very near -1, therefore we can find points of arbitrary large y-values that iterate into -1, so that the Julia set extends towards infinity in the vertical direction.

As cos(z){\displaystyle cos(z)} has power series expansion 1−z2/2!+z4/4!−z6/6!+z8/8!−...{\displaystyle 1-z^{2}/2!+z^{4}/4!-z^{6}/6!+z^{8}/8!-...} (where n! = 1×2×...×n), we can get rational approximations to the Mandelbrot set and the Julia sets by restricting this series.

The Mandelbrot set for (1−z2)/(z−z2cos(z))+c{\displaystyle (1-z^{2})/(z-z^{2}cos(z))+c}

The Julia set for (1−z2)/(z−z2cos(z))−1{\displaystyle (1-z^{2})/(z-z^{2}cos(z))-1}

The Julia set for (1−z2)/(z−z2cos(z))−1{\displaystyle (1-z^{2})/(z-z^{2}cos(z))-1}